Maximum frequency of emission is obtained for the transition:

Options

A. n = 2 to n = 1
B. n = 6 to n = 2
C. n = 1 to n = 2
D. n = 2 to n = 6

Answer

To solve the question regarding the maximum frequency of emission obtained for a transition, we can follow these steps:

### Step 1: Understand the transitions
- Emission occurs when an electron transitions from a higher energy level to a lower energy level. Conversely, absorption occurs when an electron moves from a lower energy level to a higher energy level.

### Step 2: Identify the options
- We are given four options for the transitions. We need to identify which transitions correspond to emission.

### Step 3: Eliminate absorption transitions
- From the options, eliminate any transitions that go from a lower energy level to a higher energy level, as these represent absorption.

### Step 4: Write the formula for frequency
- The frequency \( F \) of the emitted radiation can be expressed as:
\[ F = \frac{C}{\lambda} \]
where \( C \) is the speed of light and \( \lambda \) is the wavelength.

### Step 5: Use the Rydberg formula
- The Rydberg formula for the frequency of transitions is given by:
\[ F = R C z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

### Step 6: Calculate frequencies for the relevant transitions
- For the transition from \( n_2 = 2 \) to \( n_1 = 1 \):
\[ F_1 = R C z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R C z^2 \left( 1 - \frac{1}{4} \right) = R C z^2 \left( \frac{3}{4} \right) \]
Thus, we can express this as:
\[ F_1 = k \cdot \frac{3}{4} \]
where \( k = R C z^2 \).

- For the transition from \( n_2 = 6 \) to \( n_1 = 2 \):
\[ F_2 = R C z^2 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = R C z^2 \left( \frac{1}{4} - \frac{1}{36} \right) \]
Finding a common denominator (36):
\[ F_2 = R C z^2 \left( \frac{9}{36} - \frac{1}{36} \right) = R C z^2 \left( \frac{8}{36} \right) = R C z^2 \left( \frac{2}{9} \right) \]
Thus, we can express this as:
\[ F_2 = k \cdot \frac{2}{9} \]

### Step 7: Compare frequencies
- Now we compare \( F_1 \) and \( F_2 \):
- \( F_1 = k \cdot \frac{3}{4} \)
- \( F_2 = k \cdot \frac{2}{9} \)

### Step 8: Determine which frequency is greater
- To compare \( \frac{3}{4} \) and \( \frac{2}{9} \):
- Convert both to a common denominator or cross-multiply:
- \( 3 \times 9 = 27 \) and \( 2 \times 4 = 8 \)
- Since \( 27 > 8 \), we conclude that \( F_1 > F_2 \).

### Conclusion
- Therefore, the maximum frequency of emission is obtained for the transition from \( n_2 = 2 \) to \( n_1 = 1 \).

### Final Answer
- The correct option is **Option 1**.